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riadik2000 [5.3K]
3 years ago
11

If f(x) is a third degree polynomial function how many distinc complex rules are possible ?

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
5 0

Answer:

2

Step-by-step explanation:

According to the fundamental theorem of Algebra

A polynomial of degree 3 has 3 roots

However, complex roots occur in conjugate pairs, hence

There can only be 2 distinct complex roots.

You might be interested in
Correctly complete this sequence: 88511, 16351, ?, 10251
Temka [501]

Answer:

73155 is the required 3rd number that will complete the sequence as: 88511, 16351, 73155 , 10251.

Step-by-step explanation:

So, let us consider 88511 and determine the pattern to find the next number and so on to complete the sequence.

Adding the first 2 digits in the number 88511 i.e.  8 + 8 = 16

Subtracting the 3rd digit from the 2nd digit i.e. 8 - 5 = 3

Multiplying the 3rd digit and 4th digit i.e. 5 × 1 = 5

Dividing the 4th digit by the 5th digit i.e. 1 ÷ 1 = 1

Hence, using these steps would get us the number 16351.

So, the next number would be obtained using same steps:

Adding the first 2 digits in the number 16351 i.e.  1 + 6 = 7

Subtracting the 3rd digit from the 2nd digit i.e. 6 - 3 = 3

Multiplying the 3rd digit and 4th digit i.e. 3 × 5 = 15

Dividing the 4th digit by the 5th digit i.e. 5 ÷ 1 = 5

Hence, using these steps would get us the number 73155.

So, the next number would be obtained using same steps:

Adding the first 2 digits in the number 73155 i.e.  7 + 3 = 10

Subtracting the 3rd digit from the 2nd digit i.e. 3 - 1 = 2

Multiplying the 3rd digit and 4th digit i.e. 1 × 5 = 5

Dividing the 4th digit by the 5th digit i.e. 5 ÷ 5 = 1

Hence, using these steps would get us the final number 10251.

So, from this pattern method, we determine that 73155 will be the 3rd number in the sequence: 88511,16351,?,10251.

So, the complete sequence would be: 88511, 16351, 73155, 10251.

Keywords: sequence, number

Learn more about sequence from brainly.com/question/12373616

#learnwithBrainly

8 0
4 years ago
Solve the inequality <br><br> -5x &lt; 35
Marianna [84]
<span>-5x < 35
5x > -35
x > -35/5
x > -7</span>
7 0
4 years ago
As a single logarithm, 2log2 6- log29+ 1/3log2 27 = log2 12.
Ipatiy [6.2K]

Answer:

true

Step-by-step explanation:

2log2 6- log2 9+ 1/3log2 27 =

= log2 6² - log2 9 + log2 27^(1/3)

= log2 36 - log2 9 + log2 3

= log2 [(36 × 3)/9]

= log2 12

The statement is true.

3 0
3 years ago
In what different ways can 45 be written as the product of two positive integers?
nevsk [136]
First you need to know its common factors.

They are: 1, 3, 5, 9, 15, 45

And 45 = 1*45 = 3*15 = 5*9

So there should be 3 ways in total
4 0
4 years ago
How is the sum expressed in sigma notation?<br><br> 11 + 17 + 23 + 29 + 35 + 41
Advocard [28]

Answer:

The given sums 11+17+23+29+35+41  can be written as in sigma notation is  \sum\limits_{i=1}^{6}5+6i

Therefore 11+17+23+29+35+41=\sum\limits_{i=1}^{6}5+6i

Step-by-step explanation:

Given series is 11+17+23+29+35+41

To find the given sum expressed in sigma notation :

11+17+23+29+35+41  can be written as the given sums in sigma notation is  \sum\limits_{i=1}^{6}5+6i

That is

11+17+23+29+35+41=\sum\limits_{i=1}^{6}5+6i

Now expand the sums and to verify that whether the summation is true or not :

\sum\limits_{i=1}^{6}5+6i=(5+6(1))+(5+6(2))+(5+6(3))+(5+6(4))+(5+6(5))+(5+6(6))

=(5+6)+(5+12)+(5+18)+(5+24)+(5+30)+(5+36)

=11+17+23+29+35+41

Therefore 11+17+23+29+35+41=\sum\limits_{i=1}^{6}5+6i

8 0
3 years ago
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